40g of glucose (Molar mass =180) is mixed with 200‌mL of water. The freezing point of solution is ........... K. (Nearest integer) [Given, Kf=1.86K‌kg‌mol−1, density of water =1.00gcm−3, freezing point of water =273.15K]
40 g of glucose mixed with 200 mL of water 180 g of glucose = 1 moles of glucose 40 g of glucose mol = 0.22 mol 1‌mL of water =1g of water [d=1g∕cm2,1‌mL=1cm3] 200‌mL of water =200g of water ΔTf=Kfm where, ΔTf= depression in freezing point, Kf= molal elevation constant =1.86‌Kkg‌mol−1 and m = molality of solution ΔTf=